The normal to the curve x^2 + 2xy - 3y^2 = 0 | vedclass

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(A) The given curve is $x^2 + 2xy - 3y^2 = 0$.Factoring the expression: $x^2 + 3xy - xy - 3y^2 = 0 \Rightarrow x(x + 3y) - y(x + 3y) = 0 \Rightarrow (

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(A) The given curve is $x^2 + 2xy - 3y^2 = 0$.Factoring the expression: $x^2 + 3xy - xy - 3y^2 = 0 \Rightarrow x(x + 3y) - y(x + 3y) = 0 \Rightarrow (x + 3y)(x - y) = 0$.This represents two lines: $x - y = 0$ and $x + 3y = 0$.The point $(1,1)$ lies on the line $x - y = 0$. The slope of this line is $m = 1$.The normal to the curve at $(1,1)$ is perpendicular to the tangent at that point. Since the curve is a pair of lines,the normal at $(1,1)$ is perpendicular to the line $x - y = 0$.The slope of the normal is $m' = -1/m = -1$.The equation of the normal at $(1,1)$ is $(y - 1) = -1(x - 1) \Rightarrow y - 1 = -x + 1 \Rightarrow x + y = 2$.To find where this normal intersects the curve again,we find the intersection of $x + y = 2$ and the other line $x + 3y = 0$.From $x + 3y = 0$,we have $x = -3y$.Substituting into $x + y = 2$: $-3y + y = 2 \Rightarrow -2y = 2 \Rightarrow y = -1$.Then $x = -3(-1) = 3$.The intersection point is $(3, -1)$.Since the $x$-coordinate is positive and the $y$-coordinate is negative,the point $(3, -1)$ lies in the $4^{th}$ quadrant.

The normal to the curve $x^2 + 2xy - 3y^2 = 0$ at the point $(1,1)$ intersects the curve again in which quadrant?

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